A remote sensing satellite of earth revolves in a circular orbit at a height of \(0.25 \times10^6~\text{m}\) above the surface of the earth. If Earth’s radius is \(6.38\times10^6~\text{m}\) and \(g=9.8~\text{ms}^{-2}\), then the orbital speed of the satellite is:
1. \(7.76~\text{kms}^{-1}\)
2. \(8.56~\text{kms}^{-1}\)
3. \(9.13~\text{kms}^{-1}\)
4. \(6.67~\text{kms}^{-1}\)

A satellite S is moving in an elliptical orbit around the earth. If the mass of the satellite is very small as compared to the mass of the earth, then:

Kepler's third law states that the square of the period of revolution (T) of a planet around the sun, is proportional to the third power of average distance r between the sun and planet i.e. T²= Kr³, here K is constant. If the masses of the sun and planet are M and m respectively, then as per Newton's law of gravitation, the force of attraction between them is F = $\frac{\mathrm{GMm}}{{\mathrm{r}}^2}$, here G is the gravitational constant. The relation between G and K is described as:

1. GK = 4$\mathrm{\pi }$2

2. GMK = 4$\mathrm{\pi }$2

3. K = G

4. K = $\frac{\mathrm{I}}{\mathrm{G}}$

Two spherical bodies of masses \(M\) and \(5M\) and radii \(R\) and \(2R\) are released in free space with initial separation between their centres equal to \(12R.\) If they attract each other due to gravitational force only, then the distance covered by the smaller body before the collision is:

A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would Earth (mass = 5.98×10²⁴ kg) have to be compressed to be a black hole?

1. ${10}^{-9}m$

2. ${10}^{-6}m$

3. ${10}^{-2}m$

4. 100 m

Dependence of intensity of gravitational field (E) of the earth with distance (r) from the centre of the earth is correctly represented by:

 

1.		2.
3.		4.

A body of mass \(m\) is taken from the Earth’s surface to the height equal to twice the radius \((R)\) of the Earth. The change in potential energy of the body will be: